We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon. [NCERT EXEMPLAR]
We know that,
the sum of the interior angles of a polygon with 3 sides, a1 = 180°,
the sum of the interior angles of a polygon with 4 sides, a2 = 360°,
the sum of the interior angles of a polygon with 5 sides, a3 = 540°,
As, $a_{2}-a_{1}=360^{\circ}-180^{\circ}=180^{\circ}$ and $a_{3}-a_{2}=540^{\circ}-360^{\circ}=180^{\circ}$
i. e. $a_{2}-a_{1}=a_{3}-a_{2}$
So, $a_{1}, a_{2}, a_{3}, \ldots$ are in A.P.
Also, $a=180^{\circ}$ and $d=180^{\circ}$
Since, the sum of the interior angles of a 3 sided polygon $=a$
So, the sum of the interior angles of a 21 sided polygon $=a_{19}$
Now,
$a_{19}=a+(19-1) d$
$=180^{\circ}+18 \times 180^{\circ}$
$=180^{\circ}+3240^{\circ}$
$=3420^{\circ}$
So, the sum of the interior angles for a 21 sided polygon is 3420°.